Physics Ncert Solutions Class 12th

3.4 (a) Let $R_1$ = 2 Ω, $R_2$ = 4 Ω, $R_3$ = 5 Ω

If the equivalent resistance is R, then $\frac{1}{R}$ = $\frac{1}{R_1}$ + $\frac{1}{R_2}$ + $\frac{1}{R_3}$ = $\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{5}$ = $\frac{10+5+4}{20}$ = $\frac{19}{20}$

R = $\frac{20}{19}$ = 1.05 Ω

(b) The EMF of the battery = 20 V

Current through $R_{1,}$ $I_1$ = $\frac{V}{R_{1,}}$ = $\frac{20}{2}$ = 10 A

Current through $R_{2,}$ $I_2$ = $\frac{V}{R_{2,}}$ = $\frac{20}{4}$ = 5A

Current through $R_{3,}$ $I_3$ = $\frac{V}{R_{3,}}$ = $\frac{20}{5}$ = 4 A

Total current I = $I_1$ + $I_2$ + $I_3$ = 10 + 5 + 4 = 19 A

3.3 (a) The equivalent resistance of the resistor in series is given by

R = 1 + 2 + 3 = 6 Ω

(b) From Ohm's law, I = $\frac{V}{R}$ we get I = $\frac{12}{6}$ = 2 A.

Potential drop across 1 Ω resistor = I $*R$ = 2 $*1$ = 2 V

Potential drop across 2 Ω resistor = I $*R$ = 2 $*2$ = 4 V

Potential drop across 3 Ω resistor = I $*R$ = 2 $*3$ = 6 V

3.2 EMF of the battery = 10 V

Internal resistance of the battery, r = 3 Ω

Current in the circuit, I = 0.5 A

Let the resistance of the resistor be R

According to Ohm's law

I = $\frac{E}{(R+r)}$

R + r = $\frac{E}{I}$ or R = $\frac{E}{I}$ - r = $\frac{10}{0.5}$ - 3 = 17 Ω

Terminal voltage of the battery when the circuit is closed is given by

V = IR = 0.5 $*17=8.5V$

Therefore the resistance is 17 Ω and the terminal voltage is 8.5 V

3.1 EMF of the battery, E = 12 V

Internal resistance of the battery, r = 0.4 Ω

Let the maximum current drawn = I

According to OHM's law, E = Ir

So I = $\frac{E}{r}$ = $\frac{12}{0.4}$ amp = 30 amp

Therefore, the maximum current can be drawn is 30 ampere.

As per the NCERT Textbooks

“An integrated Chip consists of many passive and active components fabricated on a single chip of silicon. these ICs are compact, low-cost, and highly reliable. They consume less power and have high speed.”

These ICs are used in majority of our daily use electronics and all advance electronics. 

2.35 According to Gauss's law, the electric field between a sphere and a shell is determined by the charge $q_1$ on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge $q_2$ . For positive charge $q_1$ , potential difference V s always positive.

2.34 Equidistant planes parallel to the x-y plane are the equipotential surfaces.

Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases

Concentric spheres centered at the origin are equipotential surfaces.

A periodically varying shape near the given grid is the equipotential surface. This shape gradually reaches the shape of planes parallel to the grid at a larger distance.

2.32  Length of the co-axial cylinder, l = 15 cm = 0.15 m

Radius of the outer cylinder, $r_1$ = 1.5 cm = 0.015 m

Radius of the inner cylinder, $r_2$ = 1.4 cm = 0.014 m

Charge on the inner cylinder, q = 3.5 $\mu C=3.5*{10}^{-6}$ C

Capacitance of a co-axial cylinder of radii $r_1$ and $r_2$ is given by the relation

C = $\frac{2\pi {\epsilon }_{0}l}{{\log }_e?\frac{r_1}{r_2}}$ , where ${\epsilon }_0$ = permittivity of free space = 8.854 $*{10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

C = $\frac{2\pi *8.854\mathrm{}*{10}^{-12}*0.15}{{\log }_e?\frac{0.015}{0.014}}$ = 1.21 $*{10}^{-10}$ F

Potential difference of the inner cylinder is given by

V = $\frac{q}{C}$ = $\frac{3.5*{10}^{-6}}{1.21\mathrm{}*{10}^{-10}}$ = 2.893 $*{10}^4$ V